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sukhopar [10]
4 years ago
10

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

table (initially at rest) begins to rotate with its rate of rotation constantly increasing. 1)What is the first event that will occur
Physics
1 answer:
inn [45]4 years ago
6 0

Answer:

The answer to the question is

The ladybug begins to slide

Explanation:

To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same

Where the  frictional force equals F_{Friction} = μ×N = m×g×μ

and the centripetal force is given by m·ω²·r

If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have

m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁

and for the gentleman bug we have

m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂

But r₁ = 2×r₂

Therefore substituting the values of r₁ =2×r₂ we have

g×μ = ω²·r₁ = g×μ = ω²·2·r₂

Therefore   ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide

The ladybug begins to slide

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A moving particle fragments or decays into a particle moving at .53c, mass 135 MeV/c2, and a particle moving at .98c, mass 938 M
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Answer:

M = 1073 Mev/c2

u = 0.95 C        

Explanation:

given data:

m1 =135 Mev/c2

v1 = 0.53 c

m2 = 938 Mev/c2

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from conservation of momentum principle we have

\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = \frac{m1v1}{\sqrt{1-\frac{V1^2}{C^2}}} +\frac{m1v1}{\sqrt{1-\frac{V2^2}{C^2}}}

\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = \frac{135*0.53c}{0.848} +\frac{938*0.98c}{0.2}

\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = 4680.6 C   ...............1

Total mass of INITIAL particle M   =m1+m2 = 1073 Mev/c2

using equation 1

\frac{1073 u}{\sqrt{1-\frac{u^2}{C^2}}}= 4680.6C

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Answer:

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